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Thread: Integration with Euler's constant and cosine

  1. #1
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    Integration with Euler's constant and cosine

    Integrate

    e^(2x)cosx dx

    I tried substitution, where u=2x, but then I am still left with an x value from cosx. I know how to integrate them separately, but now when they are multipled together... how should I go about integrating?
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  2. #2
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    How's your "by parts"? Do it twice.
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  3. #3
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    Quote Originally Posted by bhuang View Post
    Integrate

    e^(2x)cosx dx

    I tried substitution, where u=2x, but then I am still left with an x value from cosx. I know how to integrate them separately, but now when they are multipled together... how should I go about integrating?
    You need to use integration by parts.

    $\displaystyle \int{u\,dv} = u\,v - \int{v\,du}$.


    So for $\displaystyle \int{e^{2x}\cos{x}\,dx}$

    Let $\displaystyle u = e^{2x}$ so that $\displaystyle du = 2e^{2x}$.

    Let $\displaystyle dv = \cos{x}$ so that $\displaystyle v = \sin{x}$.


    Therefore $\displaystyle \int{e^{2x}\cos{x}\,dx} = e^{2x}\sin{x} - \int{2e^{2x}\sin{x}\,dx}$

    $\displaystyle = e^{2x}\sin{x} - 2\int{e^{2x}\sin{x}\,dx}$.


    Now you need integration by parts again.

    Let $\displaystyle u = e^{2x}$ so that $\displaystyle du = 2e^{2x}$.

    Let $\displaystyle dv = \sin{x}$ so that $\displaystyle v = -\cos{x}$.


    So $\displaystyle e^{2x}\sin{x} - 2\int{e^{2x}\sin{x}\,dx} = e^{2x}\sin{x} - 2\left[-e^{2x}\cos{x} - \int{-2e^{2x}\cos{x}\,dx}\right]$

    $\displaystyle = e^{2x}\sin{x} + 2e^{2x}\cos{x} - 4\int{e^{2x}\cos{x}\,dx}$.



    So we have

    $\displaystyle \int{e^{2x}\cos{x}\,dx} = e^{2x}\sin{x} + 2e^{2x}\cos{x} - 4\int{e^{2x}\cos{x}\,dx}$

    $\displaystyle 5\int{e^{2x}\cos{x}\,dx} = e^{2x}\sin{x} + 2e^{2x}\cos{x}$

    $\displaystyle \int{e^{2x}\cos{x}\,dx} = \frac{1}{5}e^{2x}\sin{x} + \frac{2}{5}e^{2x}\cos{x} + C$.
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